Past exam of the perturbation methods and applied complex variables course of the University of Oxford / 2008 / A / 2

Write $y=e^{S/\delta }$ and divide the equation by $y$:Balancing the first and third terms gives $\delta =ϵ$. With $S=S_0+ϵS_1+\cdots$, the terms of order ${ϵ}^{-1}$ and one giveFor the slow branch $S_0^{\prime }=0$, choose $S_0=0$ and obtain $S_1=-x^2/2$. For the fast branch $S_0^{\prime }=-1$, choose $S_0=-x$ and obtain $S_1=x^2/2$. Additive integration constants enter the two independent amplitudes. The two-term WKB approximation is consequentlyImposing the boundary conditions on these two approximating modes yieldsThis uses the first two exponent functions $S_0,S_1$ for both branches. The next transport equations, if desired to identify the algebraic error, give $S_2=x-x^3/3$ for the slow mode and $S_2=x+x^3/3$ for the fast mode. Thus each displayed mode has relative amplitude correction $1+O(ϵ)$ uniformly on the fixed interval; no turning point occurs because the two leading exponent roots remain distinct.

Dropping the small highest-derivative term gives $y_o^{\prime }+x{y}_o=0$. The outer solution satisfying the right endpoint condition isIt fails the left endpoint condition, so a boundary layer is needed at $x=0$. If $X=x/{ϵ}^{\gamma }$, diffusion and first derivative balance when ${ϵ}^{1-2\gamma }={ϵ}^{-\gamma }$, giving $\gamma =1$.

With $y(x)=Y(X)$, the exact inner expansion after multiplication by $ϵ$ isIts leading solution is $Y=A+B{e}^{-X}$. The zero boundary value gives $B=-A$, and asymptotic matching condition as $X\to \infty$, $x\to 0$ fixes $A=e^{1/2}$. ThusThe common overlap region value is $e^{1/2}$. Add the outer and inner approximations and subtract that overlap region:The decaying inner exponential function can accommodate the left condition, whereas a right-end inner coordinate would give a mode that grows into the interval. There is no right boundary layer. The composite asymptotic expansion satisfies the left condition exactly and the right one up to the exponentially small term $e^{1/2-1/ϵ}$; the remaining approximation error is algebraic and vanishes as $ϵ\downarrow 0$.